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Let $X^{3}+AX+B$ be an irreducible abelian cubic polynomial in $Z[X]$ . We determine explicitly integers $a_{1},$ $\cdots,$ $a_{t},$ $F$ such that, exoept for finitely many primes $p$, $x^{3}+Ax+B\equiv 0(mod p)$ has three $solutions\Leftrightarrow p\equiv a_{1},$ $\cdots,$ $a_{t}(mod F)$ . Let $X^{3}+AX+B$ be an irreducible abelian cubic polynomial in $Z[X]$ . We are interested in determining those primes $p$ for which the congruence $x^{3}+Ax+B\equiv 0(mod p)$ has exactly three solutions, that is, those primes $p$ for which $X^{3}+AX+B$ splits completely into distinct linear factors modulo $p$ . As $X^{3}+AX+B$ is abelian, it is known from class field theory (see for example [6]) that, apart from a finite number ofexceptions, the primes $p$ which split $X^{3}+AX+B$ modulo $p$ lie in certain congruence classes modulo the conductor of $X^{3}+AX+B$ . In this note we determine these congruence classes expIicitly as well as the exceptional primes. Let $N_{p}(A, B)$ denote the number of solutions $x(mod p)$ of the congruence $x^{3}+Ax+B\equiv 0(mod p)$ and let $K=K(A, B)$ denote the largest positive integer such that $K^{2}|A$ and $K^{3}|B$ . Since $N_{p}(A, B)=\left\{\begin{array}{ll}N_{p}(A/K^{2}, B/K^{3}) , & if pfK,\\1 , & if p|K,\end{array}\right.$ it suffices to determine the primes $p$ for which $N_{p}(A, B)=3$ under the simplifying assumption (1) $K(A, B)=1$ . The irreducible polynomial $X^{3}+AX+B$ is abelian if and only if its discriminant is a perfect square, that is, if and only if Received May 19, 1993 * Research supported by Natural Sciences and Engineering Research Council of Canada Grant A-7233. 468 JAMES G. HUARD, BLAIR K. SPEARMAN AND KENNETH S. WILLIAMS (2) $-4A^{3}-27B^{2}=C^{2}$ for some positive integer $C$ (see [4: Example 2, p. 308]). We see from (2) that $A<0$ , $B\equiv C(mod 2)$ and $A\equiv 0$ or 2 $(mod 3)$ . Clearly $B\neq 0$ as $X^{3}+AX+B$ is irreducible. From (1) and (2) it is easy to show that exactly one of the following occurs: (3)(i) 3 $\int A$ , (3)(ii) $3\Vert A,$ $31^{\prime}B$ , (3)(iii) $3^{2}\Vert A,$ $3^{2}\Vert B$ . If (3) (i) holds then 3 $1^{\prime}C$. If (3) (ii) holds then $3^{2}|C$ and, if $3^{2}\Vert C,$ $3$ divides exactly one of $B\pm(C/9)$ . If (3) (iii) holds then $3^{3}\Vert C$ and 3 divides exactly one of $(B/9)\pm(C/27)$ . It is convenient to define the integer $b=b(A, B)=0,1,2$ by (4) $1_{b=2}^{b=0}b=1$ , $ififif3||A3||A3\int A3\int^{or}B,3^{2}||C3\int B,$ $3\Vert A,3\int B,3^{3}|C3^{2}||C’ 3|B-(C3|B+(C/9)/9)$ $oror$ $3^{2}3^{2\Vert_{A}}A,$ $3^{2}3^{2\Vert_{B}}B,$ $33|_{(B/9)-(C/27)}(B/9)+(C/27)$ , We note that (5) $b\neq 0$ $\Leftrightarrow$ $3\Vert A,$ $31^{\prime}B,$ $3^{2}\Vert C$ or $3^{2}\Vert A,$ $3^{2}\Vert B$ . In order to state our main result we need the notion of a cubic residue symbol. An Eisenstein integer $\theta$ is a complex number of the form $\theta=x+y\omega$ , where $x$ and $y$ are rational integers and $\omega=(-1+\sqrt{-3})/2$ is a complex cube root of unity. Equivalently $\theta$ is of the form $(a_{1}+a_{2}\sqrt{-3})/2$ , where $a_{1}$ and $a_{2}$ are rational integers with $a_{1}\equiv a_{2}$ $(mod 2)$ . The complex conjugate of $\theta$ is denoted by ff. The norm $N(\theta)$ of $\theta$ is the rational integer $\theta ff$. The Eisenstein integer $\theta$ is called a unit if $N(\theta)=1$ . The only units are $\pm 1$ , $\pm\omega,$ $\pm\omega^{2}$ . An Eisenstein integer $\theta$ is said to be primary if $\theta\equiv-1(mod 3)$ . For each Eisenstein integer $\theta$ not divisible by $\sqrt{-3}$ there is a unique unit $\eta=\eta(\theta)$ such that $\eta\theta$ is primary. The Eisenstein primes (up to multiplication by a unit) are $\sqrt{-3}$, rational primes of the form $3n+2$ , and Eisenstein integers with norm equal to a rational prime of the form $3n+1$ . Each nonzero Eisenstein integer can be written uniquely as a product of a unit, a nonnegative integral power of the Eisenstein prime $\sqrt{-3}$, and nonnegative integral powers of primary Eisenstein primes. If $\pi$ is an Eisenstein prime with $N(\pi)\neq 3$ , and $\theta$ is an Eisenstein integer not divisible by $\pi$ , then the cubic residue symbol $[\theta/\pi]_{3}$ is defined to be the unique cuberoot of unity such that $\theta^{(N(\pi)-1)/3}\equiv[\theta/\pi]_{3}$ $(mod \pi)$ . The basic properties of the cubic residue symbol, extended multiplicatively to denominators not divisible by $\sqrt{-3}$, are given in [3]. Before stating and proving our main result, we introduce some notation. If $a$ is a PRIMES SPLITTING ABELIAN CUBICS 469 rational integer, the integers $a^{\prime}$ and $a^{\prime\prime}$ are given uniquely by $a=3a^{\prime}+a^{\prime\prime}$ , $a^{\prime\prime}=-1,0,1$ . As usual $\phi$ denotes Euler’s phi function. We prove the following theorem. THEOREM. Let $X^{3}+AX+B\in Z[X]$ be an irreducible abelian cubic polynomial in $Z[X]$ satisfying (1). Let $C$ be the positive integer given by (2). Let $\lambda$ denote the Eisenstein integer (6) $\lambda=+(3B+C)+3B\omega=*(C+3B\sqrt{-3})$ of norm $N(\lambda)=-A^{3}$ . (i) We have (7) $(\sqrt{-3})^{c}\Vert\lambda$ , where $3^{c}\Vert A^{3}$ Let $\tau$ be the (possibly empty) product of primary Eisenstein primes such that $\lambda/((\sqrt{-3})^{c}\tau^{3})$ is cubefree. Then there is a unique product $\rho$ of primary Eisenstein primes such that (8) $ N(\rho)=\prod_{q|A,q|B}qq\langle prim\epsilon$ ) $\equiv 1\langle mod 3$ ) and $\rho N(\rho)|\lambda/((\sqrt{-3})^{c}\tau^{3})$ . (ii) With $b$ as defined in (4), we set (9) $F=3^{\alpha}N(\rho)$ , where (10) $\alpha=\left\{\begin{array}{l}0,ifb=0\\2,ifb\neq 0\end{array}\right.$ Then $F\neq 1$ and there are $\phi(F)/3$ integers $a$ satisfying (11) $1\leq a<F$ , $GCD(a, F)=1$ , $[a/\rho]_{3}=\omega^{ba’ a^{\prime\prime}}$ (iii) Let $a_{1},$ $\cdots,$ $a_{\phi\langle F)/3}$ be the $\phi(F)/3$ integers satisfying (11). Then, except for finitely many primes $p$ , we have (12) $x^{3}+Ax+B\equiv 0(mod p)$ has 3 solutions $\Leftrightarrow$ $p\equiv a_{1},$ $\cdots,$ $a_{\phi\langle F)/3}(mod F)$ . The exceptional primes are those primes $p(\neq 3)$ such that $p|C,$ $p\int F$ together with the prime 3 if $3^{4}|C$ . We note that as an exceptional prime $p$ divides $C$, it divides the discriminant of the polynomial $X^{3}+AX+B$ and so $N_{p}(A, B)\neq 3$ . Before proving this theorem we give two illustrative examples. 470 JAMES G. HUARD, BLAIR K. SPEARMAN AND KENNETH S. WILLIAMS EXAMPLE 1. We consider the irreducible abelian cubic $X^{3}-21X-17$ . Here $A=-21=-3\cdot 7,$ $B=-17$ and by (2) $C=171=3^{2}\cdot 19$ . From (4), (8), (10), (9) we see respectively that $b=1,$ $\rho=1,$ $\alpha=2,$ $F=9$ . By (11) the $\phi(F)/3=2$ integers $a_{1},$ $a_{2}$ are the solutions $a$ of $1\leq a<9$ , $GCD(a, 9)=1$ , $\omega^{aa}’=1$ . The following table shows that $a_{1}=1,$ $a_{2}=8$ . By Theorem (iii) the only exceptional prime is $p=19$ , so that for a prime $p\neq 19$ we have $x^{3}-21x-17\equiv 0(mod p)$ has3so1utions $\Leftrightarrow$ $p\equiv 1,8$ (mod9). EXAMPLE 2. We consider the irreducible abelian cubic $X^{3}-21X+35$ . Here $A=-21=-3\cdot 7,$ $B=35=5\cdot 7$ and by (2) $C=63=3^{2}\cdot 7$ . Thus from (6) we have $\lambda=\neq(63+105\sqrt{-3})$ . By (7) we see that $(\sqrt{-3})^{3}\Vert\lambda$ . Further, as $\frac{\lambda}{(\sqrt{-3})^{3}}=\frac{-35+7\sqrt{-3}}{2}=\omega^{2}(\frac{1+3\sqrt{-3}}{3})(\frac{1-3\sqrt{-3}}{2})^{2}$ , we see by (8) that $\tau=1$ and $\rho=+(1-3\sqrt{-3})$ . From (4), (10), (9) we deduce respectively $b=2,$ $\alpha=2,$ $F=3^{2}\cdot 7=63$ . By (11) the $\phi(F)/3=12$ integers $a_{1},$ $\cdots,$ $a_{12}$ are the solutions

The Primes for Which an Abelian Cubic Polynomial Splits